6.5 - Binomial and Poisson tests

You already know

For $X\sim\mathrm B(n,p)$, $\mathrm P(X=r)=\binom nrp^r(1-p)^{n-r}$. The binomial probability formula is in MF19. In a test, substitute the null value of $p$, with independent trials and a common success probability.

Worked example 1: a decrease in failure probability

Question. Historically, an independently tested component fails with probability $0.4$. After a process change, $1$ of $10$ components fails. Test at the $5\%$ level whether the failure probability has decreased, using the binomial distribution.

Plan. Use the lower tail under the historical probability, including zero failures.

Let $p$ be the failure probability after the change. Then

Under $H_0$, $X\sim\mathrm B(10,0.4)$. Thus

Since $0.0463574<0.05$, reject $H_0$.

Answer. There is sufficient evidence at the $5\%$ level that the failure probability has decreased below $0.4$.

Check. $\mathrm P(X=1)=0.040310784$ is only part of the tail; adding $\mathrm P(X=0)=0.0060466176$ gives the required result.

Worked example 2: a two-tailed investigation

Question. For the same process and observed count, test $H_0:p=0.4$ against $H_1:p\ne0.4$ at the $5\%$ level using equal tails.

Plan. Compare the lower-tail probability with $0.025$, since departures in both directions matter.

Under $H_0$, $X\sim\mathrm B(10,0.4)$, and

The observation is not in the lower rejection tail. The equal-tail critical region is actually $X=0$ or $X\ge8$, since

The adjacent lower boundary fails because $\mathrm P(X\le1)>0.025$.

Answer. Do not reject $H_0$. There is insufficient evidence at the $5\%$ level that the failure probability has changed.

Check. The two rejection tails have total probability $0.0183411712$, below $0.05$. They cannot in general be made exactly $2.5\%$ each with whole-number counts.

Exam tip

Show the probability expression, its numerical value and the comparison. A lone small probability or a single-term calculation does not justify a tail test.

You already know

For $X\sim\operatorname{Po}(\lambda)$, $\mathrm P(X=r)=e^{-\lambda}\lambda^r/r!$; this formula is in MF19. A suitable process has independent, singly occurring events at a constant rate. A longer observation interval scales the mean by its length.

Worked example 3: a one-hour count

Question. Calls to a service are modelled by a Poisson process with historical rate $4$ per hour. Investigate a suspected increase at the $5\%$ level. Find the one-hour critical region and conclude when $9$ calls are received.

Plan. Use $\operatorname{Po}(4)$ under the null and compare adjacent upper tails.

Let $\lambda$ be the mean hourly call count. Use

For a one-hour count $X$, under $H_0$,

The first exceeds $0.05$; the second does not.

Answer. The critical region is $X\ge9$, with actual significance probability $0.0214$. Nine calls lead to rejection: there is sufficient evidence at the $5\%$ level that the hourly call rate has increased.

Check. The difference between the two tails is $\mathrm P(X=8)$, about $0.0298$; including that one extra count would push the probability above $5\%$.

Worked example 4: a two-hour count

Question. For the same service, $9$ calls are received in a two-hour observation. Carry out the upper-tailed $5\%$ test of the hourly rate.

Plan. Retain $H_0:\lambda=4$, $H_1:\lambda>4$, but model the two-hour total with mean $2\lambda$.

Under $H_0$, $Y\sim\operatorname{Po}(8)$. The observed upper tail is

Do not reject $H_0$.

Answer. There is insufficient evidence at the $5\%$ level that the hourly call rate exceeds $4$.

Check. The observed rate is now $9/2=4.5$ calls per hour, much closer to the historical rate of $4$. A larger count is needed for rejection over two hours: $\mathrm P(Y\ge13)=0.0637972>0.05$ but $\mathrm P(Y\ge14)=0.0341807<0.05$.